TITLE: An alternating series identity with a hidden hyperbolic tangent
QUESTION [7 upvotes]: How does one use the inverse Mellin transform to prove that the following identity holds?
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n(e^{n\pi} + 1)} = \frac{1}{8}(\pi - 5\log(2))$$
The identity follows from MO189199 and the penultimate identity on this page (for $x=\frac{1}{2}$).

Scratch-work: I computed the Mellin transform of 
$$f(x) = \frac{(-1)^x}{x(e^{x\pi} + 1)}$$
and re-wrote the function in terms of its inverse Mellin Transform as (substituting $x = n$)
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n(e^{n\pi} + 1)} = \frac{1}{2\pi i}\int_{C} (2 \pi)^{1-s} \Gamma(s-1)\big(\zeta(s-1, \frac{1}{2} - \frac{i}{2}) - \zeta(s-1, 1-\frac{i}{2})\big)\zeta(s)ds$$
But I am not altogether confident that this has been computed correctly, and I am still not sure how to find the poles and compute the residues. For example, checking for a residue at $s=0$ gives
$$\frac{1}{8}(\pi - 2\pi i)$$
There seems to be a nontrivial contribution at $s = 1$ and $s = -1$ as well, but already the mathematics has exceeded what I understand; I am especially unclear with regard to how one accounts for residues with respect to the generalized Riemann zeta function $\zeta(s,a)$.
Clarity on using this approach or another to demonstrate the main identity would be appreciated!

REPLY [2 votes]: The following approach does not use the Mellin transform, but it is worth mentioning. 
It is quite easy to prove that:
$$\sum_{n=1}^{+\infty}\frac{\cos(\pi n)}{n} e^{-nk\pi} = -\log\left(1+e^{-k\pi}\right)$$
hence the original sum equals:
$$-\log\prod_{k=1}^{+\infty}\frac{1+e^{-(2k-1)\pi}}{1+e^{-2k\pi}}.$$
The last product is clearly related with the Jacobi triple product.
In particular, since the values of the Jacobi theta functions $\vartheta_3$ and $\vartheta_4$ in $q=e^{-\pi}$ are known, the claim simply follows.